已知5sin2α=sin2°,则的值是A.-B.C.D.2
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B
解析分析:将2a拆分成(a+1°)+(a-1°),2°拆分成(a+1°)-(a-1°),然后利用正弦函数的和角和差角公式展开,化简可得结论.
解答:5sin2a=sin2°5sin[(a+1°)+(a-1°)]=sin[(a+1°)-(a-1°)]=5sin(a+1°)cos(a-1°)+5cos(a+1°)sin(a-1°)=sin(a+1°)cos(a-1°)-cos(a+1°)sin(a-1°)∴4sin(a+1°)cos(a-1°)=-6cos(a+1°)sin(a-1°)两边除以cos(a-1°)cos(a+1°):得4tan(a+1°)=-6tan(a-1°)∴=-=-故选B.
点评:本题主要考查了两角和与差的正切函数和正弦函数,以及同角三角函数间的基本关系,属于中档题.